Question Number 146860 by mathdanisur last updated on 16/Jul/21
$$\int{sin}\left({x}\right)\:{cos}\left({x}\right)\:{dx}\:=\:? \\ $$
Answered by puissant last updated on 16/Jul/21
![=∫(1/2)sin(2x)dx =(1/2)[−(1/2)cos(2x)]+k =−(1/4)cos(2x)+k.](https://www.tinkutara.com/question/Q146861.png)
$$=\int\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\left(\mathrm{2x}\right)\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\left(\mathrm{2x}\right)\right]+\mathrm{k} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}\left(\mathrm{2x}\right)+\mathrm{k}. \\ $$
Commented by mathdanisur last updated on 16/Jul/21
$${thanks}\:{Ser} \\ $$
Answered by EDWIN88 last updated on 16/Jul/21
$$\int\mathrm{sin}\:\left(\mathrm{x}\right)\mathrm{cos}\:\left(\mathrm{x}\right)\mathrm{dx}=\int\mathrm{sin}\:\left(\mathrm{x}\right)\mathrm{d}\left(\mathrm{sin}\:\left(\mathrm{x}\right)\right) \\ $$$$=\int\:\mathrm{u}\:\mathrm{du}\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{u}^{\mathrm{2}} +\mathrm{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{c} \\ $$
Commented by mathdanisur last updated on 16/Jul/21
$${thanks}\:{Ser} \\ $$